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2C4H10 + 13O2 → 8CO2 + 10H2O This is the equation of ignition of cooking gas butane.Calculate the volume of CO2 in STP during the complete ignition of 14 kg of cooking gas. |
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Answer» Mass of Butane (C4H10) = 14 kg = 1400g GMM of C4H10 = 4 × 12 + 10 × 1 = 58 g ∴ Number of moles in molecules = \(\frac{1400}{58}\) = 241.38 Amount of CO2 when 2 moles of C4H10 ignites = 8 moles of C4H10 ignites = \(\frac{8}{2}\)× 965.52 moles ∴ Volume of CO2 formed in STP = 965.52 × 22.4 L = 21627.65 L |
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